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vesna_86 [32]
1 year ago
6

Can someone show me hwo to do this so that i know how to do this please

Mathematics
2 answers:
spayn [35]1 year ago
5 0
Y=y coordinate (2in this case)
x=x coordinate (-2 in this case)
m=slope (-2 in this case)
b= y int. (we dont know the number because it doesn’t label it so we just put the letter and you can solve for b if the question tells you so)

so the answer is
y=mx+b (slope int. form)
2=-2(-2)+b
spin [16.1K]1 year ago
3 0

Answer:

y = -2x - 2

Step-by-step explanation:

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Shawna and her best friend Keisha go shopping. The function p(t) = 3x +2x-4x2+ 21 represents how much money each girl spent base
NNADVOKAT [17]

Answer:

\$30

Step-by-step explanation:

we have

p(t)=3x+2x-4x^{2}+21

<em>Find the amount of money that each girl spent</em>

For t= 2 hours

p(2)=3(2)+2(2)-4(2)^{2}+21

p(2)=10-16+21

p(2)=\$15

<em>Find the amount of money that they spend together</em>

Multiply by 2 the amount of money that each girl spent

(2)\$15=\$30

3 0
3 years ago
6 radians is the same as what degree
olya-2409 [2.1K]
1~radian=\frac{180}{\pi}degrees\\\\1~degree=\frac{\pi}{180}radians
8 0
3 years ago
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A dilation produces a larger figure. Which is a possible scale factor? 0 Three-fourths 1 5.
laiz [17]

Answer:

5

Step-by-step explanation:

big brain and i took test

4 0
2 years ago
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The probability that a professor arrives on time is 0.8 and the probability that a student arrives on time is 0.6. Assuming thes
saul85 [17]

Answer:

a)0.08  , b)0.4  , C) i)0.84  , ii)0.56

Step-by-step explanation:

Given data

P(A) =  professor arrives on time

P(A) = 0.8

P(B) =  Student aarive on time

P(B) = 0.6

According to the question A & B are Independent  

P(A∩B) = P(A) . P(B)

Therefore  

{A}' & {B}' is also independent

{A}' = 1-0.8 = 0.2

{B}' = 1-0.6 = 0.4

part a)

Probability of both student and the professor are late

P(A'∩B') = P(A') . P(B')  (only for independent cases)

= 0.2 x 0.4

= 0.08

Part b)

The probability that the student is late given that the professor is on time

P(\frac{B'}{A}) = \frac{P(B'\cap A)}{P(A)} = \frac{0.4\times 0.8}{0.8} = 0.4

Part c)

Assume the events are not independent

Given Data

P(\frac{{A}'}{{B}'}) = 0.4

=\frac{P({A}'\cap {B}')}{P({B}')} = 0.4

P({A}'\cap {B}') = 0.4 x P({B}')

= 0.4 x 0.4 = 0.16

P({A}'\cap {B}') = 0.16

i)

The probability that at least one of them is on time

P(A\cup B) = 1- P({A}'\cap {B}')  

=  1 - 0.16 = 0.84

ii)The probability that they are both on time

P(A\cap  B) = 1 - P({A}'\cup {B}') = 1 - [P({A}')+P({B}') - P({A}'\cap {B}')]

= 1 - [0.2+0.4-0.16] = 1-0.44 = 0.56

6 0
3 years ago
PLEASE HELP ASAP 50 POINTS BRAINLIEST IF CORRECT
beks73 [17]
I think the answer is B
7 0
2 years ago
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